Recent content by ilyas.h

1. a in T, there exists n such that a^n = e b in T, there exists m such that b^m = e (ab)^mn = (a^n)^m (b^m)^n = e^m e^n = e axiom 1 holds. 2. let e be identiy in G e^n = e ==> e in T(G) T(G) subgroup of G, it's also true that ea = ae =a for all a in G conclusion: idenity of G, e...

Homework Statement Homework Equations subgroup axioms: 1. a, b in T(G), then ab in T(G) 2. existence of identity element. 3. a in T(G), then a^-1 in T(G) The Attempt at a Solution 1. let a be in T(G), then a^n = e. let b be in T(G), then b^n = e (ab)^n = (a^n)(b^n) = (e)(e) = e axiom 1...

My fault all along. I was trying to calculate the magnitude of the surface instead of the magnitude of the gradient of the surface. Yes i know how to calculate the magnitude of a vector thank you very much. answered.

how does that help? i need to know how to compute the magnitude.

nope. Take the magnitude of the llevel surface?

Homework Statement Find vector normal to z = x^2 + y^2 - 3 at point r = (2, -1, 2) Homework EquationsThe Attempt at a Solution here is the markscheme. I understand how to find the gradient, but i don't understand how they calculated the magnitude. thanks

so essentially I've done the problem but in an "all in one" fashion?

Homework Statement Let V be a vector space over a field F and let L and M be two linear transformations from V to V. Show that the subset W := {x in V : L(x) = M(x)} is a subspace of V .The Attempt at a Solution I presume it's a simple question, but it's one of those where you just don't...

can you look at my edit? thanks.

I think i understand. L(1) means that we are considering the function f(x) = 1 (a straight line through y=1): L(f) = f' + f(-2)t L(1) = d(1)/dt + f(-2)t f(-2) is equal to 1 in this case, so: d(1)/dt + f(-2)t = 0 +t = tHowever, if we consider L(t): L(t) = t' + f(-2)t =d(t)/dt + f(-2)t = 1 +...

Still doesn't make sense. L(1) = 1' + 1*(-2)t = d(1)/dt + (-2)t = 0 + (-2)t = (-2)t =/= t. what is f(-2) equal to anyway?

Homework Statement Here is the question, i know how to do part (i) but I do not understand part (ii): The Attempt at a Solution [/B] here's the solution from the marking scheme: i understand how they formed the matrix from their working out (i can se the pattern), but I do not understand...

Duh! you're awesome. Thanks.

Homework Statement [/B] find directional derivative at point (0,0) in direction u = (1, -1) for f(x,y) = x(1+y)^-1The Attempt at a Solution grad f(x,y) = ( (1+y)^-1, -x(1+y)^-2 ) grad f(0,0) = (1, 0) grad f(x,y) . u = (1,0).(1,-1) = 1. seems easy but markscheme says I am wromg. It says...

someone please help me!